Problem: Divide the following complex numbers. $ \dfrac{-7+i}{4+3i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${4-3i}$ $ \dfrac{-7+i}{4+3i} = \dfrac{-7+i}{4+3i} \cdot \dfrac{{4-3i}}{{4-3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-7+i) \cdot (4-3i)} {(4+3i) \cdot (4-3i)} = \dfrac{(-7+i) \cdot (4-3i)} {4^2 - (3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-7+i) \cdot (4-3i)} {(4)^2 - (3i)^2} = $ $ \dfrac{(-7+i) \cdot (4-3i)} {16 + 9} = $ $ \dfrac{(-7+i) \cdot (4-3i)} {25} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-7+i}) \cdot ({4-3i})} {25} = $ $ \dfrac{{-7} \cdot {4} + {1} \cdot {4 i} + {-7} \cdot {-3 i} + {1} \cdot {-3 i^2}} {25} $ Evaluate each product of two numbers. $ \dfrac{-28 + 4i + 21i - 3 i^2} {25} $ Finally, simplify the fraction. $ \dfrac{-28 + 4i + 21i + 3} {25} = \dfrac{-25 + 25i} {25} = -1+i $